Question

1) A student dissolved 0.1916 g of an unknown diprotic acid in 100 mL of distilled water. The acid was then titrated with 0.1025M NaOH solution. The second equivalence point showed the sharpest change in pH, and so it was used to determine the molar mass of the unknown acid. The volume of NaOH needed to reach the equivalence point was 27.5 mL.

a. Calculate the number of moles of NaOH used in the titration to reach the second equivalence point.



b. Calculate the number of moles of diprotic acid, based on the fact that we are examining the second equivalence point.



c. Calculate the molar mass of the diprotic acid.

Answer 1

a.
`2.818 * 10^(-3) mol\ NaOH`

b.
`1.409 * 10^(-3) mol\ acid`

c.
`135.98 (g)/(mol)`

Step-by-step explanation:

`H_2X + 2NaOH \rightarrow Na_2X +2H_2O`

a. To calculate the number of moles of NaOH we need the following definition of molarity:

`M= (mol)/(V)\\mol=M* V= 0.1025 (mol)/(L)* 27.5mL* (1L)/(1000mL)= 2.818 * 10^(-3) mol`

b. As we have a diprotic acid, it means that when the second equivalence point is reached, two moles of NaOH have reacted by one mol of acid, therefore the relation we use to determine the moles of acid is:

`(1\ mol\ acid)/(2\ mol NaOH)\\2.818 * 10^(-3) mol\ NaOH * (1\ mol\ acid)/(2\ mol NaOH)= 1.409 * 10^(-3) mol\ acid`

c. Finally, we determine the molar mass as follows:

`molar \ mass=(g\ acid)/(mol\ acid)= (0.1916\ g)/(1,409* 10^(-3) mol)= 135.98 (g)/(mol)`